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3 years ago

Assume that y varies inversely with x. If y=1/3 when x=1/2, find y when x=1/4.

Mathematics

1 answer:

LenKa [72]3 years ago

3 0

Answer:

y = $\frac{2}{3}$

Step-by-step explanation:

Given that y varies inversely with x then the equation relating them is

y = $\frac{k}{x}$ ← k is the constant of variation

To find k use the condition y = $\frac{1}{3}$ when x = $\frac{1}{2}$ , thus

$\frac{1}{3}$ = $\frac{k}{\frac{1}{2} }$ = 2k ( divide both sides by 2 )

k = $\frac{1}{6}$

y = $\frac{1}{6x}$ ← equation of variation

When x = $\frac{1}{4}$ , then

y = $\frac{1}{6(\frac{1}{4}) }$ = $\frac{1}{\frac{3}{2} }$ = $\frac{2}{3}$

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In order to raise money for holiday gifts for your family, you are planning on starting a dog walking and car wash business. You

kramer

Answer:

Part 1) see the explanation

Part 2) see the explanation

Part 3) see the explanation

Part 4) The graph in the attached figure

Part 5) see the explanation

Step-by-step explanation:

Part 1) Assign a variable to represent the number of hours that you will spend dog walking in November. Write an expression to represent the amount of money you need to earn while dog walking

Let

x ----> the number of hours spent dog walking in November

$12x=600$

solve for x

$x=50\ hours$

Part 2) Assign a variable to represent the number of hours that you will spend washing cars in November. Write an expression to represent the amount of money you need to earn while washing cars.

Let

y ----> the number of hours spent washing cars in November

$18y=600$

solve for y

$y=33.33\ hours$

Part 3) Write an algebraic model using inequalities that represents the total amount of money earned by dog walking and washing cars during the month of November.

Let

x ----> the number of hours spent dog walking in November

y ----> the number of hours spent washing cars in November

we know that

You need to earn at least $600 during the month of November

The word "at least" means " greater than or equal to"

$12x+18y\geq 600$

Part 4) Graph the algebraic model in the first quadrant only.

we have

$12x+18y\geq 600$

using a graphing tool

The solution is the shaded area

see the attached figure N 1

Part 5) Use the graph and algebraic model to answer the following:

a) Why does the graph exist only in the first quadrant?

Because the number of hours cannot be a negative number

b) Are you able to earn exactly $600? Use the solutions of the system to find possible combinations of outcomes that equal exactly $600. Where do all of the combinations occur in the graph?

Yes its possible to earn exactly $600. All the combinations of outcomes that equal exactly $600 occur in the solid line 12x+18y=600

c) Is it possible to earn more than $600? Use the solutions of the system to find possible combinations of outcomes that are greater than $600. Where do all of the combinations occur in the graph?

Yes its possible to earn more than $600. All the combinations of outcomes that are greater than $600 occur above the solid line 12x+18y=600

d) If you work for 10 hours walking dogs and 10 hours washing cars, will you have earned enough money for the holiday gifts?

we have the ordered pair (10,10)

substitute in the inequality

$12(10)+18(10)\geq 600$

$300\geq 600$ ----> is not true

therefore

You must not have earned enough money for holiday gifts.

e) Where does 10 hours walking dogs and 10 hours washing cars fall on the graph? Is this location representative of the solution to the algebraic model?

The ordered pair (10,10) fall below the solid line 12x+18y=600

This location is not representative of the solution to the algebraic model, because the solution of the model is above the solid line

f) How would the algebraic model be different if you needed to earn more than $600? Adjust your algebraic model to show that you must earn more than $600. Would the graph of the model be different from the original? Would you include the line in the solution?

we know that

$12x+18y=600$ ----> represent a linear equation that show the possible combinations of outcomes that equal exactly $600.

The solution is the set of ordered pairs belong to the solid line

$12x+18y\geq 600$ ----> represent a inequality that show the possible combinations of outcomes that are greater than or equal to $600. The solid line is included in this solution

The solution is the shaded area above the solid line (The solid line is included in the shaded area)

g) In complete sentences, explain the difference between a solid line and a dashed line when graphing an inequality. When graphing the two algebraic models, how did you determine which type of line to use?

we know that

When the inequality is of the form ≥ or ≤ we have a solid line

When the inequality is of the form > or < we have a dashed line

The difference is that in the solid line the line is included in the solution and in the dashed line the line is not include in the solution

h) How did you determine which part of the graph of the inequality to shade? What does the shaded area tell you? What does the area that is not shaded tell you?

we know that

when you have an inequality of the form

y> ax+b or y≥ax+b

The shaded area is above the line

when you have an inequality of the form

y< ax+b or y≤ax+b

The shaded area is below the line

The shaded area is the solution set of the inequality (ordered pairs that satisfy the inequality)

The area that is not shaded are the ordered pairs that are not solutions of the inequality (ordered pairs that satisfy the inequality)